If a ball is thrown in the air with an initial height of 4 feet, and if the ball remains in the air for 3.8 seconds, then accurate to the nearest foot, how high did it go? Remember, the acceleration due to gravity on Earth is $$ -32 \mathrm{ft} / \mathrm{sec}^{2} $$. [? ] feet

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To find the maximum height reached by the ball, we can use the kinematic equation for vertical motion:

$$
h(t) = h_0 + v_0 t + \frac{1}{2} a t^2
$$

where:
- $$ h(t) $$ is the height at time $$ t $$,
- $$ h_0 $$ is the initial height,
- $$ v_0 $$ is the initial velocity,
- $$ a $$ is the acceleration (which is negative due to gravity),
- $$ t $$ is the time.

Given:
- $$ h_0 = 4 $$ feet (initial height),
- $$ a = -32 $$ ft/s² (acceleration due to gravity),
- $$ t = 3.8 $$ seconds (total time in the air).

First, we need to find the initial velocity $$ v_0 $$. The ball will reach its maximum height at half of the total time in the air, which is $$ t = \frac{3.8}{2} = 1.9 $$ seconds.

At the maximum height, the velocity will be 0. We can use the following equation to find $$ v_0 $$:

$$
v = v_0 + at
$$

Setting $$ v = 0 $$ at $$ t = 1.9 $$:

$$
0 = v_0 - 32 \cdot 1.9
$$

Now, we can solve for $$ v_0 $$:

$$
v_0 = 32 \cdot 1.9
$$

Next, we will calculate $$ v_0 $$ and then use it to find the maximum height $$ h(1.9) $$. Let's perform these calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$32\times 1.9$$
- step1: Multiply the numbers:
  $$60.8$$
Calculate or simplify the expression $$ 4 + (32 * 1.9 * 1.9) + (0.5 * -32 * 1.9^2) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$4+\left(32\times 1.9\times 1.9\right)+\left(0.5\left(-32\right)\times 1.9^{2}\right)$$
- step1: Multiply the terms:
  $$4+115.52+\left(0.5\left(-32\right)\times 1.9^{2}\right)$$
- step2: Convert the expressions:
  $$4+115.52+\left(0.5\left(-32\right)\left(\frac{19}{10}\right)^{2}\right)$$
- step3: Multiply:
  $$4+115.52+\left(-\frac{1444}{25}\right)$$
- step4: Remove the parentheses:
  $$4+115.52-\frac{1444}{25}$$
- step5: Add the numbers:
  $$119.52-\frac{1444}{25}$$
- step6: Convert the expressions:
  $$\frac{2988}{25}-\frac{1444}{25}$$
- step7: Transform the expression:
  $$\frac{2988-1444}{25}$$
- step8: Subtract the numbers:
  $$\frac{1544}{25}$$
First, we calculated the initial velocity $$ v_0 $$:

$$
v_0 = 32 \cdot 1.9 = 60.8 \text{ ft/s}
$$

Next, we can find the maximum height reached by the ball at $$ t = 1.9 $$ seconds using the height equation:

$$
h(1.9) = 4 + (60.8 \cdot 1.9) + \left(\frac{1}{2} \cdot -32 \cdot (1.9)^2\right)
$$

Calculating this gives:

$$
h(1.9) = 4 + 60.8 \cdot 1.9 - 16 \cdot (1.9)^2
$$

After performing the calculations, we find:

$$
h(1.9) \approx 61.76 \text{ feet}
$$

Rounding this to the nearest foot, the maximum height reached by the ball is:

$$
\boxed{62} \text{ feet}
$$
The ball reached a maximum height of 62 feet.

If a ball is thrown in the air with an initial height of 4 feet, and if the ball remains in the air for 3.8 seconds, then accurate to the nearest foot, how high did it go? Remember, the acceleration due to gravity on Earth is \( -32 \mathrm{ft} / \mathrm{sec}^{2} \). [? ] feet

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